Question

Дана функция
$f(x) = 2 \sqrt{x} + \frac{3}{ {x}^{2} }$
Найдите f'(x), f'(1)

Answer 1

$f(x)=2\sqrt{x}+\frac{3}{x^2}=2\sqrt{x}+3x^{-2}\\\\ f'(x)=[2\sqrt{x}+3x^{-2}]'=[2\sqrt{x}]'+[3x^{-2}]'=\\\\ =2*[\sqrt{x}]'+3*[x^{-2}]'=2*\frac{1}{2\sqrt{x}}+3*(-2)*x^{-3}=\\\\ =\frac{1}{\sqrt{x}}-\frac{6}{x^{3}}.\\\\ f'(1)=\frac{1}{\sqrt{1}}-\frac{6}{1^{3}}=1-6=-5$